Unit 1 · Topic 1.4

1.4 Reference Frames and Relative Motion

Velocity depends on who's measuring it. To switch from one reference frame to another, you add or subtract the frame's velocity as a vector. Observers moving at constant velocity relative to each other disagree about velocities but always agree about accelerations, which is why Newton's laws work the same in every inertial frame.

Key terms

reference frame
relative velocity
inertial reference frame
observer
vector addition

Reference frames

A reference frame is the point of view you measure motion from, with its own origin and axes. A passenger asleep on a train is at rest in the train's frame but moving at 30 m/s in the ground's frame. Neither is wrong; motion is always relative to something.

An inertial reference frame is one that isn't accelerating. In it, an object with no net force moves at constant velocity, so Newton's first law holds. The ground is close enough to inertial for exam problems. A car that's braking or turning is not an inertial frame: loose objects inside seem to slide forward or sideways with no force pushing them.

Adding relative velocities

Use subscripts to keep track of "what, relative to what". Write $\vec{v}_{AB}$ for the velocity of A measured by B. Then:

$\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$

The inner subscripts match and cancel like a chain: (A relative to B) plus (B relative to C) gives A relative to C. Also, $\vec{v}_{BA} = -\vec{v}_{AB}$ : if you see a car approaching you at 5 m/s, the car's driver sees you approaching at 5 m/s from the other direction.

In one dimension this is adding signed numbers. A person walking forward at 1.5 m/s on a train moving at 20 m/s moves at 21.5 m/s relative to the ground; walking toward the back, at 18.5 m/s.

In two dimensions, add the vectors by components. Boats in rivers and planes in wind are the classic cases. The boat's velocity relative to the water plus the water's velocity relative to the ground gives the boat's velocity relative to the ground.

One event, two observers

A student on a train moving at a steady 20 m/s drops a ball. In the train's frame, the ball starts at rest and falls straight down. In the ground's frame, the ball starts with the train's 20 m/s forward and follows a parabola, exactly like a horizontally launched projectile.

Both observers see the ball land at the student's feet, and both measure the same fall time and the same downward acceleration, g. They disagree only about the horizontal velocity, which differs by the train's 20 m/s at every instant.

If the train were speeding up instead, the ball would land behind the student's feet as seen on the train. That's a sign the train is not an inertial frame.

Acceleration is the same in every inertial frame

If frame B moves at constant velocity relative to frame C, then $\vec{v}_{BC}$ is constant. Differentiating $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ with respect to time gives $\vec{a}_{AC} = \vec{a}_{AB}$ , because the derivative of a constant is zero.

So all inertial observers measure the same acceleration. Since they also agree on forces and masses, they agree that $\sum\vec{F} = m\vec{a}$ . That's why you can do physics on a smoothly moving train or plane just as on the ground.

Observers don't agree on everything. Velocities, kinetic energies and momenta all depend on the frame, while accelerations, forces and time intervals don't.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Crossing a river, pointed straight across
A river is 60 m wide and flows east at 3.0 m/s. A boat moves at 4.0 m/s relative to the water and points straight north, across the river. Find (a) the boat's velocity relative to the shore, (b) the time to cross and (c) how far downstream it lands.
Show the solutionHide the solution
1. Step 1: Let east be +x and north be +y. Boat relative to water: $\vec{v}_{BW} = 4.0\hat{j}$ m/s. Water relative to ground: $\vec{v}_{WG} = 3.0\hat{i}$ m/s.
2. Step 2: (a) $\vec{v}_{BG} = \vec{v}_{BW} + \vec{v}_{WG} = (3.0\hat{i} + 4.0\hat{j})$ m/s. Its magnitude is 5.0 m/s, at $\tan^{-1}(3.0/4.0) = 36.9^\circ$ east of north.
3. Step 3: (b) Only the north component carries the boat across: t = 60 m ÷ 4.0 m/s = 15 s. The current doesn't change the crossing time.
4. Step 4: (c) During those 15 s the boat drifts east at 3.0 m/s: 3.0 × 15 = 45 m downstream.
Answer: (a) 5.0 m/s at 36.9° east of north; (b) 15 s; (c) 45 m downstream.
Example 2Calculator allowed
Aiming upstream to land straight across
Same river (60 m wide, current 3.0 m/s east) and same boat (4.0 m/s relative to the water). Which way must the boat point to land directly across from where it started, and how long does that crossing take?
Show the solutionHide the solution
1. Step 1: To go straight north relative to the ground, the boat's east-west velocity relative to the ground must be zero. The boat must point partly west (upstream) at an angle θ west of north so that $4.0\sin\theta = 3.0$ .
2. Step 2: $\sin\theta = 0.75$ , so θ ≈ 48.6° west of north.
3. Step 3: Its northward ground speed is $4.0\cos\theta = \sqrt{4.0^2 - 3.0^2} = \sqrt{7} \approx 2.65$ m/s.
4. Step 4: Time = 60 ÷ 2.65 ≈ 22.7 s. The trade-off: the boat lands straight across but takes longer, because part of its velocity is used to cancel the current.
Answer: Point about 48.6° west of north (upstream); the crossing takes about 22.7 s.

Common mistakes

Adding speeds instead of velocity vectors. In two dimensions, add components; 4.0 m/s and 3.0 m/s at right angles give 5.0 m/s, not 7.0 m/s.
Getting the subscripts backward. Write $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ with the inner letters matching, and flip the sign when you flip the subscripts.
Thinking a current changes the crossing time when the boat points straight across. Perpendicular components are independent; only the cross-river component matters for time.
Assuming observers in different inertial frames see different accelerations or forces. They see different velocities but the same acceleration.

On the exam

Expect a scenario seen by two observers, such as a ball dropped on a moving train, and questions about what each one sees. Both see the same acceleration (g down), but one sees a straight line and the other a parabola.
For boat and plane problems, draw the vector triangle and label every velocity with its "relative to" subscript before calculating.

Connected topics

Videos

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Check yourself

4 questions on 1.4 Reference Frames and Relative Motion. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A train moves east at 25 m/s relative to the ground. A passenger walks toward the back of the train at 1.5 m/s relative to the train. What is the passenger's velocity relative to the ground?

Question 2 of 4Calculator allowed

Rain falls straight down at 8.0 m/s relative to the ground. A car drives through it at 6.0 m/s. What is the speed of the raindrops relative to the driver?

Question 3 of 4Calculator allowed

A passenger on a train moving at constant velocity drops a ball. The passenger and a person standing beside the track both watch the ball fall. Which quantity do they measure to be the same?

Question 4 of 4Calculator allowed

Car A drives east at 15 m/s and car B drives west at 20 m/s, toward each other on a straight road. What is the velocity of car B relative to car A?

0 of 4 answered